Question

A 5 digit no. Is made using 1,3,5,7,9 without repeat .what is sum of all such possible no.

Answer 1

We have 5 different values we're working with (1, 3, 5, 7 and 9). The number of 5-digit numbers possible without repetition of any digit is equivalent to the number of unique ways to arrange our 5 different digits (i.e. the number of 'permutations').

Here's one way to calculate this number:

For the 1st digit we can choose any of the 5 numbers, for the 2nd digit we then have 4 numbers to choose from (because we can't choose the one we've already chosen), for the 3rd digit we have 3 choices, then 2 for the 4th digit, and 1 for the 5th digit. The total unique arrangements is then: 5*4*3*2*1 (commonly written 5!) = 120 different 5-digit numbers.

Out of these 120, our 5 possible values will each appear an equal number of times in the units place, the tens place, the hundreds place, etc. In other words, we will have the same amount of numbers starting with 5 as we will numbers starting with 1, 3, 7 and 9, the same amount of numbers ending in 1 as numbers ending in 3, 5, 7 and 9, etc. This applies to every place value in our 5-digit numbers.

120/5 = 24, so there will be 24 numbers starting with 1, and 24 starting with 3, and 24 starting with 5, etc...

This helps us because clearly we don't want to actually add up all of these 120 numbers. We need a shortcut if we're going to find the sum.

We can make use of the fact that in any 5-digit number we have one digit representing 10,000, another representing 1000, another for 100, and 10 and 1. For example, in 53971, the 5 represents 5*10,000, the 3 is 3*1000, etc...

So we know each of our 5 values will appear in each position 24 times. This means that we can sum them up like this:

Sum of ten thousands digits: 24*10,000*(1 + 3 + 5 + 7 + 9)
+
Sum of thousands digits: 24*1,000*(1 + 3 + 5 + 7 + 9)
+
Sum of hundreds digits: 24*100*(1 + 3 + 5 + 7 + 9)
+
Sum of tens digits: 24*10*(1 + 3 + 5 + 7 + 9)
+
Sum of units digits: 24*1*(1 + 3 + 5 + 7 + 9)

= 6,666,600